Problem: $g(t) = t^{3}+6t^{2}-2t-4-3(f(t))$ $h(n) = 2n^{3}+5n^{2}-f(n)$ $f(t) = -3t$ $ f(g(-6)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(-6)$ . Then we'll know what to plug into the outer function. $g(-6) = (-6)^{3}+6(-6)^{2}+(-2)(-6)-4-3(f(-6))$ To solve for the value of $g$ , we need to solve for the value of $f(-6)$ $f(-6) = (-3)(-6)$ $f(-6) = 18$ That means $g(-6) = (-6)^{3}+6(-6)^{2}+(-2)(-6)-4+(-3)(18)$ $g(-6) = -46$ Now we know that $g(-6) = -46$ . Let's solve for $f(g(-6))$ , which is $f(-46)$ $f(-46) = (-3)(-46)$ $f(-46) = 138$